Question

Suppose that two conformations A and B of a molecule differ in energy by 5.0 kJ/ mol, and a third conformation lies 0.50 kJ/mol above B. What proportion of molecules will be in conformation B at 273 K, with each conformation treated as a single energy level?

Answer 1

Answer:

Let N_A = molecules in A level

Let N_B = molecules in B level

Let N_C = molecules in C level

According to Boltzmann distribution

N_B/N_A = exp(-(E_B-E_A)/RT) = exp(-5.0 x 1000J / 8.314 x 273) = 0.111

N_C/N_B = exp(-(E_C-E_B)/RT) = exp(-0.5 x 1000J / 8.314 x 273) = 0.803

Also N_A + N_B + N_C = N_A + 0.111 N_A + N_C = N_A + 0.111 N_A + 0.803 N_B =   N_A + 0.111 N_A + 0.803 x 0.111 N_A = 1.200 N_A = N_o

So No = 1.200 N_A     so N_A = (1.0/1.200) No

N_B = 0.111 N_A = 0.111 x (1.0/1.200) N_o = 0.0925 N_o

So percentage abundance of B state is = (N_B /N_o) x 100 = 0.0925 x100 = 9.25 %